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melomori [17]
3 years ago
11

If you are given the mass of an object in pounds, the time in seconds, and the distance in feet, what must you do before you can

calculate the momentum in SI units?​
Physics
1 answer:
Alexandra [31]3 years ago
5 0

Answer: First you must convert pound in kilogram, and feet in meter

Explanation:

To calculate momentum we use .

p=m*V

mass-m

speed-V

distance and time are used to calculate velocity(speed)

You are given :

 mass- in pounds

for distance - in feet

before you do any calculation first you have to convert pounds in kilograms

and feet in meters.

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Vitek1552 [10]

my bad I was in a herie last time can you please answer my question , I am going to give you the 5 points back for this question and extra 30 points ,

Explanation:

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8 0
3 years ago
The tension in the rope is constant and equal to 40 N as the block is pulled. What is the instantaneous power (in W) supplied by
Eduardwww [97]

Complete Question:

A 10 kg block is pulled across a horizontal surface by a rope that is oriented at 60° relative to the horizontal surface.

The tension in the rope is constant and equal to 40 N as the block is pulled. What is the instantaneous power (in W) supplied by the tension in the rope if the block when the block is 5 m away from its starting point? The coefficient of kinetic friction between the block and the floor is 0.2 and you may assume that the block starting at rest.

Answer:

Power = 54.07 W

Explanation:

Mass of the block = 10 kg

Angle made with the horizontal, θ = 60°

Distance covered, d = 5 m

Tension in the rope, T = 40 N

Coefficient of kinetic friction, \mu = 0.2

Let the Normal reaction = N

The weight of the block acting downwards = mg

The vertical resolution of the 40 N force, f_{y} = 40sin \theta

\sum f(y) = 0

N + 40 sin \theta - mg = 0\\N = -40sin60 + 10*9.81 = 0\\N = 63.46 N

\sum f(x) = 0

40 cos 60 - f_{r} - ma = 0\\ f_{r} = \mu N\\ f_{r} = 0.2 * 63.46\\ f_{r} = 12.69 N\\40cos 60 - 12.69-10a = 0\\7.31 = 10a\\a = 0.731 m/s^{2}

v^{2}  = u^{2} + 2as\\u = 0 m/s\\v^{2}  =  2 * 0.731 * 5\\v^{2}  = 7.31\\v = \sqrt{7.31} \\v = 2.704 m/s

Power, P = Fvcos \theta

P = 40 *2.704 cos60\\P = 54.074 W

7 0
3 years ago
A gravitational _____ exists between you and every object in the universe.
andrew-mc [135]
A pair of equal gravitational forces ... one in each direction ...
exists between every speck of mass in the universe and every
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4 0
3 years ago
Read 2 more answers
A 350-N child is in a swing that is attached to a pair of ropes 2.10 m long. Find the gravitational potential energy of the chil
o-na [289]

Answer:

a)  U = 735 J , b) U = 125.7 J , c)   U = 0 J

Explanation:

The gravitational power energy is

      U = mg y - mg y₀

The last value is a constant, for simplicity we can make it zero, if the lowest point is at the origin of the coordinate system, which in this case we will place in the lowest part

a) Rope is horizontal

The height in this case is the same length of the rope

     y = 2.10 m

    w = mg = 350 N

    U = 350 2.10

    U = 735 J

b) when the angle is 34º

     y = L - L cos 34

    y = L (1- cos34)

    y = 2.10 (1- cos 34)

    y = 0.359 m

    U = 350 0.359

    U = 125.7 J

c) in this case this point coincides with the reference system

     y = 0

     U = 0 J

4 0
3 years ago
i wanna learn a little bit of physics so if anyone could teach me a thing or two it would be much appreciated
pentagon [3]

Answer:

... message me. what do u want to know

4 0
3 years ago
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